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Solutions Manual Continuum Mechanics Lai 4th Edition
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Solutions Manual Continuum Mechanics Lai 4th Edition Complete and Comprehensive Guide

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  • November 23, 2023
  • 248
  • 2023/2024
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  • solutions manual continuum mechanics lai 4th ed
  • solutions manual continuum mechanics lai
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Solutions Manual Continuum Mechanics
Lai 4th Edition




2-1

,CHAPTER 2, PART A

2.1 Given
1 0 2 1
 Sij  = 0 1 2 and  ai  = 2 
  
   
3 0 3  3
Evaluate (a) Sii , (b) Sij Sij , (c) S ji S ji , (d) S jk Skj (e) amam , (f) Smn aman , (g) Snmaman

Ans. (a) Sii = S11 + S22 + S33 = 1 + 1 + 3 = 5 .
(b) Sij Sij = S 2 + S 2 + S 2 + S 2 + S 2 + S 2 + S 2 + S 2 + S 2 =
11 12 13 21 22 23 31 32 33
1 + 0 + 4 + 0 + 1 + 4 + 9 + 0 + 9 = 28 .
(c) S ji S ji = Sij Sij =28.
(d) S jk Skj = S1k Sk1 + S2k Sk 2 + S3k Sk 3
= S11S11 + S12 S21 + S13S31 + S21S12 + S22 S22 + S23S32 + S31S13 + S32 S23 + S33S33
= (1)(1) + ( 0 )( 0 ) + ( 2 )( 3 ) + ( 0 )( 0 ) + (1)(1) + ( 2 )( 0 ) + ( 3 )( 2 ) + ( 0 )( 2 ) + (3)(3) = 23 .
(e) amam = a12 + a22 + a23 = 1 + 4 + 9 = 14 .
(f) Smn aman = S1na1an + S2na2an + S3na3an =
S11a1a1 + S12a1a2 + S13a1a3 + S21a2a1 + S22a2a2 + S23a2a3 + S31a3a1 + S32a3a2 + S33a3a3
= (1)(1)(1) + (0)(1)(2) + (2)(1)(3) + (0)(2)(1) + (1)(2)(2) + ( 2 )( 2 )( 3 ) + (3)(3)(1)
+ ( 0 )( 3 )( 2 ) + (3)(3)(3) = 1 + 0 + 6 + 0 + 4 + 12 + 9 + 0 + 27 = 59.
(g) Snmaman = Smn aman =59.

2.2 Determine which of these equations have an identical meaning with a = Q a' .
i ij j
(a) a = Q a' , (b) a = Q a' , (c) a = a' Q .
p pm m p qp q m n mn


Ans. (a) and (c)

2.3 Given the following matrices
1 2 3 0
 ai  = 0  ,  Bij  = 0 5 1 
  
 2 0 2 1
Demonstrate the equivalence of the subscripted equations and corresponding matrix equations in
the following two problems.
(a) b = B a and b =  B  a  , (b) s = B a a and s = a Ba
T

i ij j ij i j


Ans. (a)


2-2

,bi = Bija j → b1 = B1 j a j = B11a1 + B12a2 + B13a3 = (2)(1) + ( 3 )( 0 ) + ( 0 )( 2 ) = 2
b2 = B2 ja j = B21a1 + B22a2 + B23a3 = 2, b3 = B3 ja j = B31a1 + B32a2 + B33a3 = 2 .




2-3

, 2 3 0 1 2
b = Ba = 0 5 1 0 = 2. Thus, bi = Bija j gives the same results as b = Ba

0 2 1 2  2
(b)
s = Bij aia j = B11a1a1 + B12a1a2 + B13a1a3 + +B21a2a1 + B22a2a2 + B23a2a3
+B31a3a1 + B32a3a2 + B33a3a3 = (2)(1)(1) + (3)(1)(0) + (0)(1)(2) + (0)(0)(1)
+(5)(0)(0) + (1)(0)(2) + (0)(2)(1) + (2)(2)(0) + (1)(2)(2) = 2 + 4 = 6.
2 3 0 1 2
and s = a Ba = 1 20 5 1 0 = 1 0 2 2 = 2 + 4 = 6 .

T
0
    
0 2 1  2  2

Write in indicial notation the matrix equation (a)  A = BC  , (b)  D  = B C 
T
2.4 and (c)

 E  = B C F  .
T


Ans. (a)  A = BC  → A = B C , (b) D = B C → A
T
=B C .
ij im mj ij mi mj
(c)  E  = B C F  → E
T
=B C F .
ij mi mk kj


2 2 2 2 2 2
2.5 Write in indicial notation the equation (a) s = A1 + A2 + A3 and (b)
+ + =0.
x12 x22 x32
2 2 2 2 + 2 + 2 = 0 → 2
Ans. (a) s = A1 + A2 + A3 = Ai Ai . (b) x2 x2 x2 x x = 0 .
1 2 3 i i


2.6 Given that Si j =aiaj and Sij =aiaj , where ai=Qmi am and aj =Qn jan , and Qik Qjk = ij .
Show that Sii =Sii .

Ans. Sij =QmiamQn jan =QmiQn jaman → Sii =QmiQniaman =mnaman =amam = Smm = Sii .

vi
2.7 Write ai = + v vi in long form.
t j
x j

Ans.
v
i = 1 → a = 1 + v v1 v1 v v1 v
= +v 1 +v +v 1 .
1
t j
x j t 1
x1 2
x2
3
x3
v2
i =2→a = +v v 2 = v 2 +v v 2 +v v2 + v v2 .
2
t j
x j t 1
x1 2
x2
3
x3
v3
i =3→ a = + v v3 v3 v v3 v
= +v 3 +v +v 3 .
3
t j
x j t 1
x1 2
x2
3
x3



2-4

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